We denote the $(A|B):=-{\rm tr}(AB)$ as the inner product of the skew-hermitian matrix, which is easy to prove by the definition. I'm stuck by the problem when I read the Lie algebra of matrix group, finding it's a calculation problem of matrix .
$A\in{{\rm SU}(2)}$
So $A$ can be written as
$$A=\left(\begin{array}{1}u&v\\-\overline{v}&\overline{u} \end{array}\right)$$
for $u,v\in\mathbb{C}$ and $|u|^2+|v|^2=1$.
So we express A in the form
$$A=\cos{\theta}I+S$$
where S is skew hermitian and ${\rm Re} \,u=\cos{\theta}$ with $\theta\in[0,\pi]$
Then we can caculate that
$S^2=-\sin^2\theta{I}$ and $(S|S)=2\sin^2\theta$
We know that $\mathfrak{su}(2)$ is all skew-hermitian matrix $\subset{M_2(\mathbb{C})}$
So the question is
If $U \in \mathfrak{su}(2)$ with $(S|U)=0$, how to prove the result $SUS=\frac{(S|S)}{2}U$?
The author in the book gives the hint that we can use the properties of the vector product, but it's hard for me to understand what he says. So I want to try a direct way to caculate the result. I guess $SU=-US$, so$SUS=-USS=-U\sin^2\theta{I}=-\sin^2\theta{U}=\frac{(S|S)}{2}U$
I denote $S$ as $\left(\begin{array}{1} \sin\theta{i}&v\\-\overline{v}&-\sin\theta{i} \end{array} \right)$ and denote $U$ as $\left(\begin{array}{1} ai&u\\-\overline{u}&bi \end{array} \right)$, but it seems wrong to prove the conjecture.
lemma:If $U_1,U_2\in\mathfrak{su}(2)$ with $U_1=x_1\hat{H}+y_1\hat{E}+z_1\hat{F}$ and $U_2=x_2\hat{H}+y_2\hat{E}+z_2\hat{F}$($\hat{H},\hat{E},\hat{F}$ form an orthonormal basis), so $$U_1U_2=-\frac{(U_1|U_2)}{2}I+\frac{1}{2}[U_1,U_2]$$ So $$SUS=\frac{1}{2}[S,U]S\Rightarrow{}SUS=\frac{1}{2}SUS-\frac{1}{2}USS\Rightarrow{}SUS=-USS=\frac{(S|S)}{2}U$$